Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.

Question

Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.

I have shown it is not possible for $a=4$, but not for any a greater than $4$.

Answer 1

We see that

$$\frac{x+y}{xy}=\frac{2^a}{7^a}$$ and that leads to $$7^a(x+y)=2^a(xy)$$ now let $\gcd(x,y)=d$, and write $x=dx_0$ and $y=dy_0$. We get: $$7^a(dx_0+dy_0)=2^a(dx_0dy_0)$$ or, rewritten: $$7^a(x_0+y_0)=2^a(dx_0y_0)$$ note that since $\gcd(x_0,y_0)=1$, we know $\gcd(x_0y_0,x_0+y_0)=1$; thus, $x_0$ and $y_0$ can only contain factors $7$ (since $x_0y_0|7^a(x_0+y_0)$ implies $x_0y_0|7^a$). Now write $x=dx_0=d\cdot7^n$ and $y=dy_0=d\cdot7^m$. We substitute this into the equation to get

$$7^a(7^n+7^m)=2^a(d\cdot7^{n+m})$$ meaning, $2^a|7^n+7^m$. Now set w.l.o.g. $n\geq m$, so that we can write $7^n+7^m=7^m(7^{n-m}+1)$. We know there's no factors $2$ in $7^m$, so we can say $2^a|7^{n-m}+1$. We can write $n-m=k$ to simplify this to $2^a|7^k+1$. Proving that $7^k+1$ can never contain four factors $2$ would thus be sufficient.

Let's look at this $\mod 16$. We see that $7^2=49\equiv 1\mod 16$, and as such, we know that $7^k+1\equiv 2\mod 16$ if $k$ is even, and if $k$ is odd, $7^k+1=7\cdot 7^{k-1}+1\equiv 7\cdot 1+1\equiv 8\mod 16$. Therefore, $7^k+1$ is never divisible by $16$, and as a result of that, never contains 4 or more prime factors. This proves that the initial equality has no integer solutions if $a>3$.

Answer 2

Rewrite the equation into

$$7^a\cdot(x+y)=2^a\cdot xy$$

From this, we conclude that there is some positive integer $k$ with

\begin{align} xy&=k\cdot 7^a\tag{1}\label{eq1}\\ x+y&=k\cdot 2^a\tag{2}\label{eq2} \end{align}

Now, from equation $\eqref{eq1}$, we conclude that if $p\neq 7$ is prime and $p|k$, then $p|x$ or $p|y$.
On the other hand, equation $\eqref{eq2}$ implies that if $p\neq2$ is prime and $p$ divides both $x$ and $k$, then $p|y$, and similarly exchanging the roles of $x$ and $y$.

Together, this means that any prime $p\neq 2,7$ in the factorization of $k$ also factors both $x$ and $y$. Say $p$ is one such prime and let $e_x$ be the exponent of $p$ in the factorization of $x$, and similarly for $e_y$. From equation $\eqref{eq1}$, $p$ has an exponent of $e_x+e_y$ in the factorization of $k$, but from equation $2$ that exponent is $\max\{e_x,e_y\}$.

It follows that no prime $p\neq 2,7$ is in the factorization of $k$, so $k,x$ and $y$ are all of the form $2^s7^t$. We may then rewrite this system. In the obvious notation, equation $\eqref{eq1}$ yields

\begin{align} s_x+s_y&=s_k\\ t_x+t_y&=t_k+a \end{align}

Equation $\eqref{eq2}$ then yields

$$2^{s_x}7^{t_x}+2^{s_y}7^{t_y}=2^{s_k+a}7^{t_k}$$

We may assume without loss of generality that $s_x\geq s_y$. Then:

$$2^{s_x-s_y}7^{t_x}+7^{t_y}=2^{s_k-s_y+a}7^{t_ k}=2^{s_x+a}7^{t_ k}$$

so if $s_x>s_y$, the RHS is odd while the LHS is odd. It follows that $s_x=s_y$ and the equation may be simplified to

$$7^{t_x}+7^{t_y}=2^{s_x+a}7^{t_k}$$

Now, we may assume without loss of generality that $t_x\geq t_y$. Then

$$7^{t_x-t_y}+1=2^{s_x+a}7^{t_k-t_y}=2^{s_x+a}7^{t_x-a}$$

We must then have $t_x=a$, for otherwise the RHS is either fractional, or a multiple of $7$ (while the LHS is not). This then implies $t_y=t_k$. Hence

$$7^{a-t_k}+1=2^{s_x+a}$$

If either $a$ were even or $s_x$ were positive and even, we could factor $2^{s_x+a}-1$ as

$$\Big({\left(2^{s_x}\right)}^{a/2}-1\Big)\Big({\left(2^{s_x}\right)}^{a/2}+1\Big)$$

Since these factors differ by $2$, at most one can be a multiple a $7$, and this would imply that the smallest is $1$, so that $\frac{as_x}2=1$. Since $a\geq 4$, this is impossible, and $a$ must be odd and $s_x$ must be either odd or $0$.

If $s_x$ were odd, then $a+s_x$ is even, and we may repeat the factorization argument:

$$\Big(2^{(s_x+a)/2}-1\Big)\Big(2^{(s_x+a)/2}+1\Big)$$

Once again $\frac{s_x+a}2=1$ cannot be true for $a\geq 4$, so $s_x$ must be $0$. We then have

$$7^{a-t_k}=2^a-1,$$

where $a\geq 5$ is odd. By looking into the equaiton above modulo $4$, we find that $t_k$ is even. Then, with $a=2\beta+1$ and $t_k=2\sigma$ we may rewrite the equation as

$$7\cdot {\left(7^2\right)}^{\beta-\sigma}=2^{2\beta+1}-1$$

Finally, looking at the equation above modulo $16$, we find that the LHS is congruent to $7$ while the RHS is congruent to $-1\equiv 15$. This completes the proof that when $a\geq 4$, no solution exists. $\square$